An algorithmic view: the Seidel triangle Bernoulli number








seidel s algorithm in fact more general (see exposition of dominique dumont (dumont 1981)) , rediscovered several times thereafter.


similar seidel s approach d. e. knuth , t. j. buckholtz (knuth & buckholtz 1967) gave recurrence equation numbers t2n , recommended method computing b2n , e2n ‘on electronic computers using simple operations on integers’.


v. i. arnold rediscovered seidel s algorithm in (arnold 1991) , later millar, sloane , young popularized seidel s algorithm under name boustrophedon transform.


triangular form:







only  a000657, 1 1, ,  a214267, 2 1s, in oeis.


distribution supplementary 1 , 1 0 in following rows:







this  a239005, signed version of  a008280. main andiagonal  a122045. main diagonal  a155585. central column  a099023. row sums: 1, 1, −2, −5, 16, 61…. see  a163747. see array beginning 1, 1, 0, −2, 0, 16, 0 below.


the akiyama–tanigawa algorithm applied  a046978 (n + 1) /  a016116(n) yields:







1. first column  a122045. binomial transform leads to:







the first row of array  a155585. absolute values of increasing antidiagonals  a008280. sum of antidiagonals − a163747 (n + 1).


2. second column 1 1 −1 −5 5 61 −61 −1385 1385…. binomial transform yields:







the first row of array 1 2 2 −4 −16 32 272 544 −7936 15872 353792 −707584…. absolute values of second bisection double of absolute values of first bisection.


consider akiyama-tanigawa algorithm applied  a046978 (n) / ( a158780 (n + 1) = abs( a117575 (n)) + 1 = 1, 2, 2, 3/2, 1, 3/4, 3/4, 7/8, 1, 17/16, 17/16, 33/32….







the first column absolute values  a000111 numerator of trigonometric function.


 a163747 autosequence of first kind (the main diagonal  a000004). corresponding array is:







the first 2 upper diagonals −1 3 −24 402… = (−1) ×  a002832. sum of antidiagonals 0 −2 0 10… = 2 ×  a122045(n + 1).


− a163982 autosequence of second kind, instance  a164555 /  a027642. hence array:







the main diagonal, here 2 −2 8 −92…, double of first upper one, here  a099023. sum of antidiagonals 2 0 −4 0… = 2 ×  a155585(n + 1). note  a163747 −  a163982 = 2 ×  a122045.







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